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Revista de la Unión Matemática Argentina

versão impressa ISSN 0041-6932

Rev. Unión Mat. Argent. vol.50 no.1 Bahía Blanca jun. 2009

 

The subvariety of Q-Heyting algebras generated by chains

Laura A. Rueda

Abstract. The variety QH of Heyting algebras with a quantifier [14] corresponds to the algebraic study of the modal intuitionistic propositional calculus without the necessity operator. This paper is concerned with the subvariety C of QH generated by chains. We prove that this subvariety is characterized within QH by the equations ∇ (x ∧ y) ≈ ∇x ∧ ∇y and (x → y) ∨ (y → x) ≈ 1 . We investigate free objects in C .

1. Introduction and Preliminaries

Distributive lattices with a quantifier were considered as algebras for the first time by Cignoli in [7] who studied them under the name of Q -distributive lattices. A Q -distributive lattice is an algebra (L; ∨,∧, 0,1,∇ ) of type (2, 2,0,0,1) such that (L; ∨,∧, 0,1) is a bounded distributive lattice and the unary operation ∇ satisfies the following conditions, for any a , b ∈ L :  ∇0 = 0 ,  a ∧ ∇a = a ,  ∇ (a ∧ ∇b ) = ∇a ∧ ∇b  and  ∇ (a ∨ b) = ∇a ∨ ∇b .  These conditions were introduced by Halmos [9] as an algebraic counterpart of the logical notion of an existential quantifier.

Various further investigations have been carried out since [7] (see R. Cignoli [8], H. Priestley [13], M. Adams and W. Dziobiak [4], M. Abad and J. P. Díaz Varela [2] and A. Petrovich [11]). As a natural generalization, the operation of quantification was considered for Heyting algebras in [3] and [15]. A Heyting algebra is an algebra (H; ∨, ∧,→, 0,1) of type (2,2,2,0,0) for which (H; ∨, ∧,0,1) is a bounded distributive lattice and for a , b ∈ H , a → b is the relative pseudocomplement of a with respect to b , i.e.,  a ∧ c ≤ b if and only if c ≤ a → b . It is known that the class of Heyting algebras forms a variety. An important subvariety of Heyting algebras is the class of linear Heyting algebras [5]. A linear Heyting algebra is a Heyting algebra that satisfies the equation (x → y) ∨ (y → x) ≈ 1 . Throughout this paper H will denote the category of Heyting algebras and Heyting algebra homomorphisms and HL will denote the subcategory of linear Heyting algebras.

A Q -Heyting algebra is an algebra (H; ∇ ) such that H is an object of H and ∇ is a quantifier on H , that is, ∇ is a unary operation defined as for Q -distributive lattices. Monadic Boolean algebras are the simplest examples of Q -Heyting algebras. The class of Q -Heyting algebras forms a variety, which we denote QH . The subvariety of QH characterized within QH by the equation (x → y) ∨ (y → x ) ≈ 1 , that is, the subvariety of linear Q -Heyting algebras will be denoted by QHL . Q -Heyting algebras were first introduced in [14] and have been investigated in [14153].

In this paper we investigate the subvariety C of the variety of Q -Heyting algebras generated by chains. We characterize C by identities in Section 2 and we investigate free objects in this variety in Section 3.

We will usually use the same notation for a variety and for the algebraic category associated with it. And, similarly we will use the same notation for a structure and for its universe.

Recall that Heyting algebras are algebraic models of the intuitionistic propositional logic and that the study of extensions of Intuitionistic Propositional Calculus (IPC) reduces to the study of subvarieties of the variety H . The language of intuitionistic modal logic (MIPC) is the language of IPC enriched with two modal unary operators of necessity    and of posibility  ◊ . The algebraic models of MIPC are the monadic Heyting algebras.

Now, in MIPC the operators    and  ◊ are independent from each other, that is p ↔ ¬ ◊¬p and ◊p ↔ ¬ ¬p are not theorems in MIPC. Hence, the set of theorems of the propositional calculus without of the necessity operator  , called the  -free fragment of MIPC is different from that of MIPC. Similarly, the set of theorems of the propositional calculus without of the possibility operator ◊ , called the ◊ -free fragment of MIPC is different from that of MIPC.

It turns out that the behaviour of the ◊ -free fragment of MIPC is very much similar to that of MIPC. However, surprisingly enough, the  -free fragment of MIPC behaves pretty diferent from MIPC.

Q -Heyting algebras are the algebraic models of the  -free fragment of MIPC, that is, Q -Heyting algebras are the  -free reducts of monadic Heyting algebras.

For a poset X and Y ⊆ X , let (Y] = {u : u ≤ v for some v ∈ Y } and [Y ) = {u : u ≥ v for some v ∈ Y } . We write [u) , (u] instead of [{u}) , ({u} ] respectively. We say that Y is decreasing if Y = (Y ] , increasing if Y = [Y ) and convex if Y = (Y ] ∩ [Y ) . A mapping φ is order preserving if φ(u) ≤ φ (v) whenever u ≤ v .

In order to describe the dual category of QH we recall that a Priestley space is a triple (X; ≤, τ) such that (X; ≤ ) is a partially ordered set, (X; τ ) is a compact topological space, and the triple is totally order-disconnected (that is, for u , v ∈ X , if u ⁄≤ v then there exists a clopen increasing U ⊆ X such that u ∈ U and v ∕∈ U ). Priestley showed that the category of bounded distributive lattices and lattice homomorphisms is dually equivalent to the category of Priestley spaces and order preserving continuous functions (see the survey paper [12]).

A Q -Heyting space  (X; E ) (see [71415]) is a Priestley space (X; ≤, τ) together with an equivalence relation E defined on X such that  (i)  (Y ] is clopen for every convex clopen Y ⊆ X , (ii)  ∇EU ∈ D (X ) for each U ∈ D (X ) , where ∇ U = {v : vEu for some u ∈ U} E and D (X ) is the lattice of clopen increasing subsets of X , and  (iii)  the blocks of E are closed in X . For a ∈ H , let σ (a) ⊆ X = X (H ) denote the clopen increasing set that represents a , where X (H ) is the set of prime filters of H , ordered by set inclusion and with the topology having as a sub-basis the sets σ (a ) = {P ∈ X (H ) : a ∈ P } and X (H ) \ σ (a ) for a ∈ H . If a, b ∈ H then, under the duality, a → b corresponds to the clopen increasing set X \ (σ (a ) \ σ(b)] .

For Q -Heyting spaces (X; E) and (Y ;E′) , a Q -Heyting morphism is a continuous order-preserving mapping φ : X → Y such that φ ([u)) = [φ(u)) and  -1 -1 ∇E φ (V) = φ (∇E ′V) , for each V ∈ D (Y ) .

It can be proved in the usual way that the category of Q -Heyting algebras and homomorphisms is dually equivalent to the category of Q -Heyting spaces and Q -Heyting morphisms [1415]. For each Q -Heyting algebra (H; ∇ ) the corresponding Q -Heyting space is (X (H );E ∇) , where E = E ∇ = {(P, Q) ∈ X (H )2 : P ∩ ∇ (H ) = Q ∩ ∇ (H )} . Conversely, if (X; E ) is a Q -Heyting space, the corresponding Q -Heyting algebra is (D (X );∇E ) , where ∇E is defined as in (ii) .

2. The variety C

In this section we will study the subvariety C generated by chains within QH . Observe that if (H; ∇) ∈ C , then H ∈ HL , that is, C ⊆ QHL ⊆ QH . Consequently, C |= (x → y) ∨ (y → x) ≈ 1 .

Recall that in the variety of Heyting algebras, congruences are determined by filters. Precisely, if H ∈ H and F is a filter of H , then θF = {(a,b) ∈ H × H : (a → b) ∧ (b → a) ∈ F } is a congruence on H , and the correspondence F ↦→ θF establishes an isomorphism from the lattice of filters of H on Con H H , the lattice of congruences of H . If F is generated by an element a , F = [a) , we write θa = θ[a) .

Observe that if C is a Heyting chain and F is a filter of C , (a,b) ∈ θF if and only if a = b or a,b ∈ F . Then, ConQH (C; ∇ ) = ConH C . As a consequence of this, we have that if C is a chain, (C; ∇ ) is a subdirectly irreducible  algebra in QH if and only if C is a subdirectly irreducible  algebra in H , that is, C has a unique dual atom.

A quantifier ∇ on an algebra H ∈ QH is said to be multiplicative if ∇ (a ∧ b) = ∇a ∧ ∇b , for every a,b ∈ H .

Let M be the subvariety of QH characterized by the equation ∇ (x ∧ y) ≈ ∇x ∧ ∇y .

Lemma 2.1. If (H; ∇) ∈ M , ConQH (H; ∇ ) = ConH H .

Proof Let θF ∈ ConH H and (a,b) ∈ θF , i.e., (a → b) ∧ (b → a ) ∈ F . As a ∧ (a → b) = a ∧ b , then ∇a ∧ ∇ (a → b) ≤ ∇b , that is, ∇ (a → b) ≤ ∇a → ∇b . So (a → b) ∧ (b → a) ≤ ∇ ((a → b) ∧ (b → a)) = ∇ (a → b) ∧ ∇(b → a) ≤ (∇a → ∇b ) ∧ (∇b → ∇a ) . Thus (∇a → ∇b ) ∧ (∇b → ∇a ) ∈ F , so (∇a, ∇b ) ∈ θ F . Therefore, θF ∈ ConQH (H; ∇) . ■

Observe that C |= {∇ (x ∧ y) ≈ ∇x ∧ ∇y, (x → y) ∨ (y → x) ≈ 1} , that is, C ⊆ M ∩ QH L . Let us see that C = M ∩ QH L .

Lemma 2.2. Let (H; ∇ ) be a subdirectly irreducible  algebra in M ∩ QHL . Then H is a chain.

Proof Let (H; ∇ ) ∈ M ∩ QHL be a subdirectly irreducible algebra. Then ConQH (H; ∇ ) = ConH H and hence H is subdirectly irreducible in H , that is, H has a unique dual atom. Since for every a,b ∈ H , (a → b) ∨ (b → a) = 1 , then a → b = 1 or b → a = 1 , that is, a ≤ b or b ≤ a . So H is a chain. ■

Corollary 2.3. C = M ∩ QHL .

As a consequence of this corollary we have that C is characterized within QH by the identities ∇ (x ∧ y ) ≈ ∇x ∧ ∇y and (x → y ) ∨ (y → x) ≈ 1 .

The following theorem characterizes the dual space of an algebra in M .

Theorem 2.4. Let (H; ∇ ) be a Q -Heyting algebra, let (X (H );E) be the associated Q -Heyting space and {Ei}i∈I the partition of X (H ) determined by E . Then, (H; ∇ ) ∈ M if and only if each E i has exactly one maximal element.

Proof Suppose that (H; ∇) ∈ M and there exists i0 ∈ I such that Ei0 has two maximal elements M , N , M ⁄= N . Let a ∈ H be such that M ∈ σ (a ) and N ∕∈ σ(a) . For each P ∈ σ(a) ∩ Ei0 , we have that N ⁄⊆ P . Thus there exists bP ∈ H such that N ∈ σ(bP ) and J ∈ X \ σ(b ) P . Consequently

 ⋃ σ(a) ∩ Ei0 ⊆ X \ σ (bP). P∈σ(a)∩Ei0

As σ (a ) ∩ Ei0 is closed, by a compacteness argument

 n n ⋃ ∧ σ(a) ∩ Ei0 ⊆ X \ σ(bi) = X \ σ( bi) = X \ σ(b) i=1 i=1

and N ∈ σ (b) . So M ∈ σ(a) ∩ Ei0 and N ∈ σ(b) ∩ Ei0 . This implies that Ei0 ⊆ σ (∇a ∧ ∇b ) and consequently σ(∇a ∧ ∇b ) ∩ Ei0 = Ei0 . On the other hand, ∇E σ(a ∧ b) ∩ Ei0 = ∇E (σ(a) ∩ σ(b)) ∩ Ei0 = ∅ , which contradicts that (H; ∇) ∈ M .

Conversely, we know that D(X (H )) ∈ QH . Let us see that ∇E σ(a) ∩ ∇E σ (b) = ∇E σ(a ∧ b) . Since ∇E is a quantifier, ∇E σ(a ∧ b) ⊆ ∇E σ (a) ∩ ∇E σ(b) . Let us prove the other inclusion. Let P ∈ ∇E σ (a) ∩ ∇E σ(b) and i0 ∈ I such that P ∈ Ei0 . Since σ (a ) ∩ Ei0 ⁄= ∅ and σ(b) ∩ Ei0 ⁄= ∅ , if {Mi0 } = max Ei0 , then M ∈ σ(a) ∩ σ (b) = σ(a ∧ b) i0 . Therefore E ⊆ ∇ σ(a ∧ b) i0 E and so P ∈ ∇E σ (a ∧ b). ■

Lemma 2.5. M is the greatest subvariety of QH such that every filter determines a congruence.

Proof Let (H; ∇ ) ∈ QH such that (H; ∇ ) ∕∈ M . We are going to construct a filter in H which does not determine a congruence. From (H; ∇ ) ∕∈ M , there exist a,b ∈ H such that ∇a ∧ ∇b ⁄≤ ∇(a ∧ b) . Then there exists a prime ideal M such that ∇ (a ∧ b) ∈ M and ∇a ∧ ∇b ⁄∈ M . Since M is an ideal we have that ∇a ∈∕M and ∇b ∕∈ M . Consider the filter F = [a) . Then (a ∧ b,b) ∈ θ F , being that ((a ∧ b) → b) ∧ (b → (a ∧ b)) = b → a ≥ a . Let us see that (∇ (a ∧ b),∇b ) ∕∈ θF . Suppose on the contrary that (∇ (a ∧ b),∇b ) ∈ θF . Thus ∇b → ∇ (a ∧ b) ≥ a , which implies that ∇b → ∇ (a ∧ b) ≥ ∇a (*) since the image of ∇ is closed under implication. On the other hand, ∇b ∧ (∇b → ∇ (a ∧ b)) ≤ ∇ (a ∧ b) , so ∇b ∧ (∇b → ∇ (a ∧ b)) ∈ M . Since M is a prime ideal and ∇b ∈∕ M we have that ∇b → ∇ (a ∧ b) ∈ M . This, together with (*), implies that ∇a ∈ M , which is a contradiction. ■

3. Free algebras

In this section we characterize the free algebra in C with n generators. Following a path analogous to that of M. Abad and L. Monteiro in [1], we will provide a method to construct the order set Π (n) of all join-irreducible elements of the free algebra, and as a consequence, we will obtain a formula to compute |Π (n)| .

It is clear that for any subset X of a chain (C; ∇ ) , the subalgebra of (C; ∇ ) generated by X is S(X ) = X ∪ ∇(X ) ∪ {0,1} . Thus, every n -generated subalgebra of a chain of C has at most 2n + 2 elements, that is, the class of all chains in C is uniformly locally finite. So C is generated by a uniformly locally finite class, and consequently, C is a variety locally finite [6, Theorem 3.7].

If (H; ∇ ) ∈ C is a finite algebra, the Q -Heyting space (X (H );≤,τ, E) has the discrete topology and (X (H );≤ ) is anti-isomorphic to the ordered set (Π(H );≤ ) of join-irreducible elements of H . In this section we will use the set Π (H ) instead of X (H ) and we will consider the relation E defined on Π (H ) , that is we consider (Π (H );E ) . If {Ei }i∈I is the partition determined by E in Π(H ) , we say that Ei ≤ Ej if and only if min Ei ≤ min Ej . This is an order relation.

Theorem 3.1. [10] A Heyting algebra is linear if and only if the family of prime filters which contain a prime filter is a chain.

Definition 3.2. Let (H; ∇ ) ∈ C be a finite algebra. Let p ∈ Π(H ) and let E1 ≤ ⋅⋅⋅ ≤ Er+1 , such that (p] ∩ Ej ⁄= ∅ , 1 ≤ j ≤ r + 1 , where (p] = {q ∈ Π(H ) : q ≤ p} . We say that p has coordinates (m, m1, ...,mr+1 ) , if the chain (p] is of length m + 1 and if mj = |(p] ∩ Ej| , 1 ≤ j ≤ r + 1 .

Notice that the set (p] of the previous definition is considered within Π (H ) .

Let m be a non negative integer. Let C = {0,a ,...,a ,1} m 1 m be the chain with m + 2 elements. Let ∇ (Cm ) = {b0 = 0,b1,...,br,br+1 = 1} , r ≤ m , with bi < bj for i < j . Let (bi,bj] be the interval in Cm consisting of the elements a ∈ Cm such that bi < a ≤ bj . We denote Cm,m1,...,mr+1 the algebra (Cm; ∇ ) , where mi = |(bi- 1,bi]| , i = 1,...,r + 1 .

Observe that if p ∈ Π (H ) , H ∕θp is a chain. More precisely, (p] is of length m + 1 if and only if H ∕θp is a chain with m + 2 elements [1, p. 7]. If π : H → H ∕θp is the natural homomorphism and a1,...,am ∈ H are such that π(0) < π (a1) < ⋅⋅⋅ < π(am ) < π(1) , then there exist join-irreducible elements q ,...,q 1 m in H such that q < ⋅⋅⋅ < q < p 1 m and π(qi) = π (ai) , 1 ≤ i ≤ m . Moreover, taking into account that qiEqj en Π(H ) if and only if ∇qi = ∇qj en H , it follows that p ∈ Π (H ) has coordinates (m, m1, , ...,mr+1) if and only if H ∕θp = Cm,m1,...,mr+1 . Since ∩p ∈Π(H) θp is the trivial relation, we have that (H; ∇ ) is a subdirect product of the chains {H ∕θp}p∈Π(H) .

Let L(n) be the free C -algebra with a finite set of generators of cardinal n > 0 . For the sake of simplicity we will write Π(n ) instead of Π (L(n )) .

We know that every n -generated subalgebra of a chain of C has at most 2n + 2 elements. Since L(n )∕θp is a chain generated by at most n elements, we have the following

Lemma 3.3. If p ∈ Π (n) , then |L(n )∕ θp| ≤ 2 n + 2 .

If p ∈ Π(n ) then from Lemma 3.3, p has coordinates (m, m1, ...,mr+1 ) , for some m , 0 ≤ m ≤ 2n and m1, ...,mr+1 ∈ IN such that ∑r+1 m = m + 1 j=1 j .

Consider the following sets:

M1 = {bj : |(bj- 1,bj]| = 1,1 ≤ j ≤ r}

and

N = Cm,m1,...,mr+1 \ ∇ (Cm,m1,...,mr+1).

For a subset T of Cm,m1,...,mr+1 to generate the algebra Cm,m1,...,mr+1 , every non constant element must be contained in T , that is, N ⊆ T . Besides, every constant can be obtained from N , except the constants of M1 . So we have that T ⊇ N ∪ M1 and consequently, |T | ≥ |N | + |M | = m - r + |M | = m - (r - |M |) 1 1 1 .

For every m , 0 ≤ m ≤ 2n , consider the sets

 r∑+1 Nm (n) = {(m1, ...,mr+1 ) : mj ∈ IN, 1 ≤ j ≤ r+1, mj = m+1, m - r+ |M1 | ≤ n }. j=1

We will denote Nm instead of Nm (n) . Observe that N0 = {(1)} , N1 = {(2),(1,1)} and N2n = {(2,2, ...,2,1)} for every n ∈ IN , that is, N2n consists of one (n + 1) -tuple whose n first coordinates are equal to 2. Moreover, if n ≥ 2 , we have that N = {(3),(2,1),(1,2),(1,1,1 )} 2 , but if n = 1 , N2 = {(2,1 )} .

Let Πm,m1,...,mr+1(n) = {p ∈ Π(n ) : L (n)∕θp = Cm,m1,...,mr+1} . It is clear that

 2⋃n ⋃ Π(n) = Πm,m1,...,mr+1(n) m=0 (m1,...,mr+1)∈Nm

and that  ′ ′ ′ Πm,m1,...,ml(n) ∩ Πm ,m1,...,m l′(n) = ∅  for   ′ ′ (m1, ...,ml) ⁄= (m 1,...,m l′) .

Let Fm,m1,...,mr+1 (n ) be the set of all functions f from the set G of free generators of L (n) into Cm,m1,...,mr+1 such that S (f(G )) = Cm,m1,...,mr+1 . Observe that every Fm,m1,...,mr+1(n) is nonempty, as f ∈ Fm,m1,...,mr+1(n) if and only if M1 ∪ N ⊆ f(G ) , that is m - r + |M | ≤ n 1 .

Recall that a filter P in a finite Heyting algebra is prime if and only if P = [p) , where p is join-irreducible element.

If f ∈ Fm,m1,...,mr+1(n) , f can be extended to a unique homomorphism -- f from L (n ) onto Cm,m1,...,mr+1 . If  -- -- N (f ) = {a ∈ L (n ) : f(a) = 1} is the kernel of -- f , it is well known that  -- N (f ) is a prime filter in L (n) , so  -- N (f ) = [pf) , with pf ∈ Πm,m1,...,mr+1 . Thus, for each (m1, ...,mr+1 ) ∈ Nm, 0 ≤ m ≤ 2n , we have a function

ψ : F (n) → Π (n) m,m1,...,mr+1 m,m1,...,mr+1 m,m1,...,mr+1

defined by ψm,m1,...,mr+1(f) = pf .

Lemma 3.4. The following holds |Πm,m1,...,mr+1(n)| = |Fm,m1,...,mr+1(n)|, 0 ≤ m ≤ 2n, (m1, ...,mr+1 ) ∈ Nm .

Proof Let us see that ψm,m ,...,m 1 r+1 is onto. For p ∈ Πm,m ,...,m (n ) 1 r+1 , consider h the natural homomorphism from L (n ) onto L (n)∕θp = Cm,m1,...,mr+1 , and f = h↾G the restriction of h to G . Then S (f(G)) = S (h(G )) = h (L (n)) = Cm,m1,...,mr+1 and therefore f ∈ Fm,m1,...,mr+1(n) . Let -- f be the extension of f . Since -- f↾G= f = h↾G , then -- f = h and therefore ψ (f) = p = p m,m1,...,mr+1 f .

Let us prove that the function ψm,m1,...,mr+1 , is one-to-one. If the functions f1 , f2 ∈ Fm,m1,...,mr+1(n ) satisfy  --- --- N (f1) = N (f2 ) then there is an automorphism α of Cm,m1,...,mr+1 such that  --- --- α ∘ f1 = f2 . But the only automorphism of C m,m1,...,mr+1 is the identity, then f-= f-- 1 2 and then f1 = f2 . ■

Lemma 3.5.

 
 2∑n ∑ |Π(n )| = |Π (n )| m,m1,...,mr+1 m=20n (m1,...,mr+1)∈Nm ∑ ∑ = |Fm,m1,...,mr+1(n)|. m=0 (m1,...,mr+1)∈Nm

If N S(a,b) is the number of functions from a set with a elements onto a set with b elements, then:

N S (a,b)  { ∑b -1 i(b) a = i=0(- 1 ) i (b - i) if a ≥ b . 0 if a < b

Let l = r - |M1 | . Then, for each (m1, ...,mr+1 ) ∈ Nm, 0 ≤ m ≤ 2n ,

 l∑+2 (l + 2) |Fm,m1,...,mr+1(n )| = N S(n,m - l + k). k=0 k

In particular, for F0,1(n) , m - l = 0 and then |F0,1(n)| = N S(n,0) + 2N S (n,1) + N S(n,2 ) = 2n . For F (n) 1,2 , m = 1 , r = |M | = 0 1 and for F (n) 1,1,1 , m = r = |M1 | = 1 . So in both cases, m - l = 1 , then  n n |F1,2(n)| = |F1,1,1(n )| = N S (n,1) + 2N S(n, 2) + N S (n,3) = 3 - 2 . And for F2n,2,2,...,2,1(n) , m = 2n , r = n , |M1 | = 0 , then m - l = n and |F2n,2,2,...,2,1(n)| = n!.

Consequently,

|Π(1)| = |F0,1(1)|+ |F1,2(1)|+ |F1,1,1(1)|+ |F2,2,1(1)| = 2 + 1+ 1 + 1 = 5.

Consider the set

 2n ⋃ ⋃ F (n) = Fm,m1,...,mr+1(n ). m=0 (m1,...,mr+1)∈Nm

If f ∈ F (n) , there is a unique m and a unique (m1, ...,mr+1 ) ∈ Nm such that f ∈ F (n) m,m1,...,mr+1 . If we put ψ(f) = ψ (f) m,m1,...,mr+1 we have a one-to-one mapping from F(n ) onto Π(n ) .

The following lemma is immediate (recall that (pf] = {q ∈ Π (n ) : q ≤ pf} ).

Lemma 3.6. pf ∈ Π(n ) has coordinates (0,1) if and only if f (g ) ∈ {0, 1} for all g ∈ G .

As a consequence, the set Π (n) has 2n minimal elements.

Lemma 3.7. For  1 ≤ m ≤ 2n , and (m1, ...,mr+1 ) ∈ Nm , pf ∈ Π (n)  has  coordinates  (m, m1, ...,mr+1 ) if and only if N ∪ M1 ⊆ f(G ) ⊆ Cm,m1,...,mr+1 .

Proof From the proof of Lemma 3.4, pf has coordinates (m, m1, ...,mr+1 ) if and only if f ∈ Fm,m1,...,mr+1 (n ) , and from the comment preceding that lemma, this is equivalent to M ∪ N ⊆ f(G ) ⊆ C 1 m,m1,...,mr+1 . ■

Remark 3.8. We know that if f ∈ Fm,m ,...,m (n) 1 r+1 , the extension homomorphism -- f and the natural homomorphism h from L (n) into  -- L (n)∕N (f ) satisfy  -- h = f . Then if in Π (n) , (pf] = {p1,...,pm, pm+1 = pf } , we have

 ( -- -- { 1 if x ∈ N (f ) f(x) = ai if x ∈ [pi) \ [pi+1), 1 ≤ i ≤ m . ( 0 if x ∕∈ [p1)

The proof of the following lemma will be omitted since it is an adaptation of that of [1, Lemma 3.13].

We say q covers p if p < q and p ≤ r < q implies r = p .

Lemma 3.9. If p , q ∈ Π (n) , q covers p if and only if the following conditions hold:

  • (i)  p < q ,

  • (ii)  p ∈ Πm,m1,...,mr+1 (n ) , 0 ≤ m ≤ 2n - 1 , (m ,...,m ) ∈ N 1 r+1 m .

  • (iii)  q ∈ Πm+1,m1,...,mr+1+1 (n ) or q ∈ Πm+1,m1,...,mr+1,1(n ) , 0 ≤ m ≤ 2n - 1 , (m1, ...,mr+1 + 1) ∈ Nm+1 and (m1, ...,mr+1, 1) ∈ Nm+1 .

In the following theorem we denote a0 = 0 .

Theorem 3.10. Let f , h ∈ F (n) . Then ψ (h) = ph covers ψ(f ) = pf if and only if f ∈ Fm,m1,...,mr+1(n) , h ∈ Fm+1,m1,...,mr+1+1(n) or h ∈ Fm+1,m1,...,mr+1,1(n ) , 0 ≤ m ≤ 2n - 1 , (m1, ...,mr+1 ) ∈ Nm , and for g ∈ G the following conditions hold:

  • (I)  f (g) = ai if and only if h (g) = ai , 0 ≤ i ≤ m .

  • (II)  f (g) = 1 if and only if h(g) = 1 or h (g) = am+1 .

Proof Suppose that ph covers pf . The first part of the theorem is an immediate consequence of Lemma 3.9

Since in Π (n) , p1 < ⋅⋅⋅ < pm < pf < ph , we have

 

-- -- f (x ) = h(x) = 0 if and only if x ∕∈ [p1),

 

-- -- f (x ) = h(x) = ai, 1 ≤ i ≤ m if and only if x ∈ [pi) \ [pi+1), 1 ≤ i ≤ m,

 

-- -- f (x ) = h(x) = 1 if and only if x ∈ [ph),

 

-- f (x ) = 1 and -- h (x ) = am+1 if and only if x ∈ [pf) \ [ph).

In particular, we have the conditions (I) and (II) .

Conversely, let f , h ∈ F (n ) be such that

f ∈ Fm,m1,...,mr+1(n), h ∈ Fm+1,m1,...,mr+1+1(n) ∪ Fm+1,m1,...,mr+1,1(n ),

and satisfying (I ) and (II) . Then,

pf ∈ Πm,m ,...,m (n) and ph ∈ Πm+1,m ,...,m +1(n )∪ Πm+1,m ,...,m ,1(n ). 1 r+1 1 r+1 1 r+1

From Lemma 3.9 , we must prove that pf < ph .

Consider in Π (n)

p1 < ⋅⋅⋅ < pm < pm+1 = pf

and

q1 < ⋅⋅⋅ < qm < qm+1 < qm+2 = ph

the chains (pf] and (ph] respectively and consider the following sets:

pict

Then

pict

We have that Cm+2 is a filter, C0 is an ideal and Ci , 0 ≤ i ≤ m , are nonempty sets, being that  -- ai ∈ h(L (n)) ,  -- ai ∈ f (L(n)) , 0 ≤ i ≤ m . Cm+1 is also nonempty. Indeed, if h ∈ Fm+1,m1,...,mr+1+1 (n ) , since f ∈ Fm,m ,...,m (n) 1 r+1 , there is g ∈ G such that f(g) ⁄= h(g) , then from (I) and (II) , -- h(g) = am+1 and -- f (g) = 1 , that is g ∈ Cm+1 . If h ∈ Fm+1,m1,...,mr+1,1(n ) , there is g ∈ G such that ∇f (g ) ⁄= ∇h (g) , then from (I) and (II) , -- h(∇g ) = am+1 and -- f(∇g ) = 1 , that is ∇g ∈ Cm+1 .

It is clear that the sets Ci , 0 ≤ i ≤ m + 2 , are pairwise disjoint. Observe that Cm+2 ∪ Cm+1 = [qm+1 ) ∩ [pm+1 ) , and so it is a filter. Using these remarks it is a routine matter to show that the set  m+2 S = ∪i=0 Ci is a subalgebra of L (n ) .

Let us see that G ⊆ S . If g ∈ G , h(g) ∈ {0,a1,...,am, am+1, 1} .

If h(g) = 1 , g ∈ [qm+2) and from (II) , f(g ) = 1 , that is, g ∈ [p ) m+1 . Then g ∈ C ⊆ S m+2 .

If h(g) = am+1 , g ∈ [qm+1 ) \ [qm+2 ) and from (II ) , f (g) = 1 , that is g ∈ [pm+1 ) . So g ∈ Cm+1 ⊆ S .

If h(g) = ai , 0 ≤ i ≤ m , then g ∈ [qi) \ [qi+1) and from (I) , f (g ) = ai , that is, g ∈ [pi) \ [pi+1) . Then g ∈ Ci ⊆ S .

Therefore, G ⊆ S and consequently S = L (n ) .

Then we can write, [p ) = [q ) ∩ L (n) = [q ) ∩ (∪m+2 C ) = ∪m+2 ([q ) ∩ C ) = [q ) ∩ C = [q ) ∩ [p ) = [p ) ∩ [p ) h m+2 m+2 i=0 i i=0 m+2 i m+2 m+2 m+2 m+1 h f .

Since ph ⁄= pf , we have pf < ph. ■

The previous theorem allows us to construct the ordered set of join-irreducible elements of the free algebra L (n) . By virtue of Lemma 3.6, there exists a one-to-one correspondence between the set of minimal elements of Π(n ) and the set of functions f from G into {0, 1} . Since (p ] is a chain, for every p ∈ Π (n) , then (f] is also a chain for f ∈ F (n ) . So, the ordered-connected components of F (n) are [f) , where f is minimal, that is, the order-connected components of F (n) are the sets [f ) , where f : G → {0, 1} .

We have constructed the Q -Heyting space (Π(n);E ) . The free algebra L (n) ∈ C with a finite set of generators of cardinality n > 0 , is the algebra obtained from (Π(n);E ) considering the decreasing subsets of Π(n ) with the quantifier given by ∇EU = {q ∈ Π (n) : qEp f or same p ∈ U } , for each U decreasing set of Π (n) .

Example 3.11. In the next figure we give the free algebra (L (1);∇ ) generated by an element g , and the ordered set Π (1) of its join-irreducible elements, with the equivalence relation which determines the quantifier.

PICT

Where 1 = (0) ∈ F0,1(1 ) , 2 = (1) ∈ F0,1(1) , 3 = (a1 ) ∈ F1,2(1) , 4 = (a1) ∈ F1,1,1(1) and 5 = (a1) ∈ F1,2,1(1) . We denote ¬g = g → 0 .

In the rest of this section we investigate the poset Π (n) in order to obtain a recursive formula for the number of elements of Π (n ) .

Let Kj(n) be the family of order-connected components [f) , with f minimal, such that |f- 1(1)| = j , 0 ≤ j ≤ n . It is clear that |K0(n )| = 1 , and if K (n ) = {K } 0 , then |K | = 1 . In general, |K (n)| = (n) j j .

For a given j , all the order-connected components in Kj (n) have the same number of elements. So if  { } Kj(n ) = K1, K2, ...,K (nj) and N (n,j ) = |K | for K ∈ Kj (n) , then

|| n || ||(⋃j) || (n ) | Ki| = N (n,j). ||i=1 || j

We are going to determine N (n, j) .

Consider K = [f) ∈ Kj(n) . From Theorem 3.10, we know that h ∈ [f ) covers f if and only if h ∈ F1,2(n ) or h ∈ F1,1,1(n ) and

  • (I)  h (g) = 0 if and only if f (g) = 0 .

  • (II)  h (g) ∈ {a1, 1} if and only if f(g) = 1 .

In particular, there are (j) (j) j 1 + ⋅⋅⋅ + j = 2 - 1 funtions h in F1,2(n) covering f , and similarly there are 2j - 1 funtions h in F1,1,1(n) covering f . So, there are 2(2j - 1) functions h covering f ,  ( ) 2 j t of which satisfy |h -1(a )| = t 1 , 1 ≤ t ≤ j .

In these conditions we have the following result.

Proposition 3.12.

  • (1)  If h ∈ F (n ) 1,1,1 , there exists f ∈ F (n) 1 0,1 with [f ) ∈ K (n) 1 j-t such that [h) and [f1) are order-isomorphic.

  • (2)  If h ∈ F1,2(n ) and h1 covers h , with h1 ∈ F2,2,1(n) and h1(g) = h(g) , for every g ∈ G , then there exists f1 ∈ F0,1(n) , [f1) ∈ Kj -t(n) such that [h ) \ [h1) , [h1) and [f ) 1 are order-isomorphic.

Proof

  • (1)  If f1 : G → {0,1} is the function defined by:

     { 1 if h(g) = 1 (*) f1(g) = 0 if h(g) = a1 or h(g) = 0 ,

    f1 is clearly a minimal element of F (n) , f1 ∈ F0,1(n) and [f1) ∈ Kj- t(n ) .

    Let us see that [h) and [f1) are order-isomorphic. Observe that if u ∈ [h) , then u ∈ F1+i,1,m2,...,mr+1(n) , where 0 ≤ i ≤ 2(j - t) and 1 ≤ r ≤ j - t + 1 . We define α : [h ) → [f ) 1 by means of α (u) = v , where

     ( { 0 if u(g) = 0 (**) v(g) = ( 1 if u(g) = 1 ak-1 if u(g) = ak 1 ≤ k ≤ 1 + i , v ∈ Fi,m2,...,mr+1(n ) .

    Clearly α is an isomorphism.

  • (2)  Observe that, if u ∈ [h) \ [h1) , u ⁄= h , then u ∈ F1+i,m1,...,mr+1(n) , 1 ≤ i ≤ 2 (j - t) , m1 ≥ 2 and 0 ≤ r ≤ j - t . If f1 is the function defined by (*) , then [h) \ [h1) and [f1) are order-isomorphic. Indeed, if we define α : [h) \ [h1) → [f1) by means of α(u) = v , where u ∈ [h) \ [h ) 1 and v defined as in (** ) , v ∈ Fi,m1-1,...,mr+1 (n ) and it can be proved that α is an isomorphism.

    Finally, consider β : [h1) → [f1) defined by β (u ) = v , where u ∈ [h1 ) , u ∈ F2+i,2,m2,...,mr+1 (n ) , 0 ≤ i ≤ 2(j - t) , 1 ≤ r ≤ j - t + 1 and

     ( { 0 if u(g) = 0 or u(g) = a1 v(g) = 1 if u(g) = 1 ( ak-2 if u(g) = ak, 2 ≤ k ≤ i + 2 , v ∈ Fi,m2,...,mr+1(n ).

    Clearly β is an isomorphism.

■

From the previous proposition,

 ∑j [ (j ) ] N (n,j) = 3 N (n, j - t) + 1. t=1 t

Therefore

||n || [ ] |(⋃j) | (n ) ∑j [ (j ) ] || Ki ||= 3 N (n, j - t) + 1 , ||i=1 || j t=1 t

and then

 [ n ( ) [ j [ ( ) ] ] ] ∑ n ∑ j |F (n)| = |Π (n)| = 1 + j 3 t N (n, j - t) + 1 . j=1 t=1

Acknowledgment: I gratefully acknowledge helpful comments of the referees. In particular, one of them suggested the algebraic proof of Lemma 2.5.

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Laura A. Rueda
Departamento de Matemática,
Universidad Nacional del Sur
Bahía Blanca, Argentina
larueda@criba.edu.ar

Recibido: 12 de noviembre de 2007
Aceptado: 2 de junio de 2009